cpp-flashcards/org/study_deck_02/dsa/graphs/0684-redundant-connection.org

#+PROPERTY: STUDY_DECK_02
* TODO 0684. Redundant Connection :medium:
:PROPERTIES:
:NEETCODE: [[file:../../roadmap.org::*0684. Redundant Connection][0684. Redundant Connection]]
:END:

In this problem, a tree is an *undirected graph* that is connected and has no cycles.

You are given a graph that started as a tree with ~n~ nodes labeled from ~1~ to ~n~, with one additional edge added. The added edge has two *different* vertices chosen from ~1~ to ~n~, and was not an edge that already existed. The graph is represented as an array ~edges~ of length ~n~ where ~edges[i] = [a_{i}, b_{i}]~ indicates that there is an edge between nodes ~a_{i}~ and ~b_{i}~ in the graph.

Return /an edge that can be removed so that the resulting graph is a tree of /~n~/ nodes/. If there are multiple answers, return the answer that occurs last in the input.

*Example 1:*


#+begin_src
Input: edges = [[1,2],[1,3],[2,3]]
Output: [2,3]
#+end_src


*Example 2:*


#+begin_src
Input: edges = [[1,2],[2,3],[3,4],[1,4],[1,5]]
Output: [1,4]
#+end_src


*Constraints:*

 - ~n == edges.length~

 - ~3 <= n <= 1000~

 - ~edges[i].length == 2~

 - ~1 <= a_{i} < b_{i} <= edges.length~

 - ~a_{i} != b_{i}~

 - There are no repeated edges.

 - The given graph is connected.

** TODO Approach
Write your approach here.

** TODO Python
#+begin_src python
class Solution:
    def findRedundantConnection(self, edges: List[List[int]]) -> List[int]:
#+end_src

** TODO C++
#+begin_src cpp
class Solution {
public:
    vector<int> findRedundantConnection(vector<vector<int>>& edges) {
        
    }
};
#+end_src